How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $sec α = - 1.108$ $secalpha=-1.108$ ?

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How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $sec α = - 1.108$ $secalpha=-1.108$ ?

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Answer 1 · 2017-02-18T03:49:46

$205^{\circ} 84$ $205^@84$

Explanation:

$sec a = \frac{1}{cos a} = - 1.108$ $sec a = 1/(cos a) = - 1.108$
$cos a = - \frac{1}{1.108} = - 0.90$ $cos a = - 1/1.108 = - 0.90$
cos a = - 0.90 -->calculator gives: arc $a = 154^{\circ} 16$ $a = 154^@16$
Unit circle gives another arc x that has the same cos value:
x = - 154^@16 or
x = 360 - 154.16 = 205^@84.
This arc lies in Quadrant 3.

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How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $sec α = - 1.108$ $secalpha=-1.108$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find the angle αalpha such that the angle lies in quadrant III and secα=−1.108secalpha=-1.108?

1 Answer

Explanation:

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How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $sec α = - 1.108$ $secalpha=-1.108$ ?